Solve Using a Matrix by Elimination 3x-7y=25 5x-8y=27

$3 x - 7 y = 25$ $5 x - 8 y = 27$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 3 & - 7 & 25 \\ 5 & - 8 & 27 \end{array}]$

Step 2

Find the reduced row echelon form of the matrix.

Perform the row operation $R_{1} = \frac{1}{3} R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $1$ .

Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = \frac{1}{3} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} \frac{1}{3} R_{1} & \frac{1}{3} R_{1} & \frac{1}{3} R_{1} \\ 5 & - 8 & 27 \end{array}]$

$R_{1} = \frac{1}{3} R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{1}{3} R_{1}$ .

$[\begin{array}{c} (\frac{1}{3}) \cdot (3) & (\frac{1}{3}) \cdot (- 7) & (\frac{1}{3}) \cdot (25) \\ 5 & - 8 & 27 \end{array}]$

$R_{1} = \frac{1}{3} R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & - \frac{7}{3} & \frac{25}{3} \\ 5 & - 8 & 27 \end{array}]$

Perform the row operation $R_{2} = - 5 \cdot R_{1} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - 5 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & - \frac{7}{3} & \frac{25}{3} \\ - 5 \cdot R_{1} + R_{2} & - 5 \cdot R_{1} + R_{2} & - 5 \cdot R_{1} + R_{2} \end{array}]$

$R_{2} = - 5 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 5 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & - \frac{7}{3} & \frac{25}{3} \\ (- 5) \cdot (1) + 5 & (- 5) \cdot (- \frac{7}{3}) - 8 & (- 5) \cdot (\frac{25}{3}) + 27 \end{array}]$

$R_{2} = - 5 \cdot R_{1} + R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & - \frac{7}{3} & \frac{25}{3} \\ 0 & \frac{11}{3} & - \frac{44}{3} \end{array}]$

Perform the row operation $R_{2} = \frac{3}{11} R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $1$ .

Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = \frac{3}{11} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & - \frac{7}{3} & \frac{25}{3} \\ \frac{3}{11} R_{2} & \frac{3}{11} R_{2} & \frac{3}{11} R_{2} \end{array}]$

$R_{2} = \frac{3}{11} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{3}{11} R_{2}$ .

$[\begin{array}{c} 1 & - \frac{7}{3} & \frac{25}{3} \\ (\frac{3}{11}) \cdot (0) & (\frac{3}{11}) \cdot (\frac{11}{3}) & (\frac{3}{11}) \cdot (- \frac{44}{3}) \end{array}]$

$R_{2} = \frac{3}{11} R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & - \frac{7}{3} & \frac{25}{3} \\ 0 & 1 & - 4 \end{array}]$

Perform the row operation $R_{1} = \frac{7}{3} R_{2} + R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $0$ .

Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = \frac{7}{3} R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} \frac{7}{3} R_{2} + R_{1} & \frac{7}{3} R_{2} + R_{1} & \frac{7}{3} R_{2} + R_{1} \\ 0 & 1 & - 4 \end{array}]$

$R_{1} = \frac{7}{3} R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{7}{3} R_{2} + R_{1}$ .

$[\begin{array}{c} (\frac{7}{3}) \cdot (0) + 1 & (\frac{7}{3}) \cdot (1) - \frac{7}{3} & (\frac{7}{3}) \cdot (- 4) + \frac{25}{3} \\ 0 & 1 & - 4 \end{array}]$

$R_{1} = \frac{7}{3} R_{2} + R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & 0 & - 1 \\ 0 & 1 & - 4 \end{array}]$

Step 3

Use the result matrix to declare the final solutions to the system of equations.

$x = - 1$

$y = - 4$

Step 4

The solution is the set of ordered pairs that makes the system true.

$(- 1, - 4)$

Solve Using a Matrix by Elimination 3x-7y=25 5x-8y=27

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